PROGRAMMING:Roche limit
An important plot in the science fiction movie wandering earth is that when the earth is too close to Jupiter, the atmosphere begins to be sucked away by Jupiter. As the earth approaches the "rigid body Roche limit", the earth is in danger of being completely torn apart. But in fact, this calculation is wrong.
Roche limit is the distance between the gravitational force of one celestial body and the tidal force caused by the second celestial body. When the distance between two celestial bodies is less than the Roche limit, the celestial body tends to break up and become the ring of the second celestial body. It was named after Edward Roche, the first person to calculate the limit（ From Baidu Encyclopedia)
The ratio of the density of a large object to that of a small object to the third power, and then multiplied by the radius of the large object and a multiple (the multiple corresponding to the fluid is 2.455, and the multiple corresponding to the rigid body is 1.26), is the value of the Roche limit. For example, the density ratio of Jupiter to the earth to the third power is 0.622. If we assume that the earth is a fluid, then the Roche limit is $$0.622 times 2.455 = 1.52701 $$times the radius of Jupiter; But the earth is a rigid body, and the corresponding Roche limit is $$0.622 times 1.26 = 0.78372 $$times the radius of Jupiter. This distance is smaller than the radius of Jupiter, that is, the earth will be torn up only when it is inside Jupiter. In other words, the earth cannot be torn up.
This question asks you to judge whether a small celestial body will be torn up by a large celestial body.
###Input format:
Input three numbers in one line, which are: the value calculated after the ratio of the density of large celestial body to that of small celestial body is 3 power ($$); the attribute of small celestial body (0 for fluid and 1 for rigid body); the ratio of the distance between two celestial bodies to the radius of large celestial body ($> 1 $$) but not more than 10).
###Output format:
In one row, the ratio of the Roche limit of small celestial body to the radius of large celestial body is first output (2 decimal places after the decimal point); After that, one space is left blank; Final output`^_^` If small objects will not be torn up, otherwise output ` t_ T`。
###Input sample 1:
```in
0.622 0 1.4
```
###Output sample 1:
```out
1.53 T_ T
```
###Input example 2:
```in
0.622 1 1.4
```
###Output example 2:
```out
0.78 ^_^
```
answer：If there is no answer, please comment
Roche limit is the distance between the gravitational force of one celestial body and the tidal force caused by the second celestial body. When the distance between two celestial bodies is less than the Roche limit, the celestial body tends to break up and become the ring of the second celestial body. It was named after Edward Roche, the first person to calculate the limit（ From Baidu Encyclopedia)
The ratio of the density of a large object to that of a small object to the third power, and then multiplied by the radius of the large object and a multiple (the multiple corresponding to the fluid is 2.455, and the multiple corresponding to the rigid body is 1.26), is the value of the Roche limit. For example, the density ratio of Jupiter to the earth to the third power is 0.622. If we assume that the earth is a fluid, then the Roche limit is $$0.622 times 2.455 = 1.52701 $$times the radius of Jupiter; But the earth is a rigid body, and the corresponding Roche limit is $$0.622 times 1.26 = 0.78372 $$times the radius of Jupiter. This distance is smaller than the radius of Jupiter, that is, the earth will be torn up only when it is inside Jupiter. In other words, the earth cannot be torn up.
This question asks you to judge whether a small celestial body will be torn up by a large celestial body.
###Input format:
Input three numbers in one line, which are: the value calculated after the ratio of the density of large celestial body to that of small celestial body is 3 power ($$); the attribute of small celestial body (0 for fluid and 1 for rigid body); the ratio of the distance between two celestial bodies to the radius of large celestial body ($> 1 $$) but not more than 10).
###Output format:
In one row, the ratio of the Roche limit of small celestial body to the radius of large celestial body is first output (2 decimal places after the decimal point); After that, one space is left blank; Final output`^_^` If small objects will not be torn up, otherwise output ` t_ T`。
###Input sample 1:
```in
0.622 0 1.4
```
###Output sample 1:
```out
1.53 T_ T
```
###Input example 2:
```in
0.622 1 1.4
```
###Output example 2:
```out
0.78 ^_^
```
answer：If there is no answer, please comment