Description

Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.

The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.

Input

The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.

Output

Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.

Sample Input

Input

4 2 1 3

Output

TRIANGLE

Input

7 2 2 4

Output

SEGMENT

Input

3 5 9 1

Output

IMPOSSIBLE

题意：

给出四个整数如果存在三个数能构成一个三角形，则输出TRIANGLE，如果任何三个数都不能构成三角形，但是存在三个数可构成退化三角形，则输出SEGMENT，如果任何三个数既不能构成三角形也不能构成退化三角形，则输出IMPOSSIBLE

退化三角形：三条边，不可以组成三角形，但是存在两条边的和等于第三条边；

AC代码：

1 #include
      
        2 #include
       
         3  4 using namespace std; 5 int dp[6]={
    0}; 6 int s[4]={
    0}; 7  8 int gon() 9 {10     if(s[1]+s[2]>s[3]&&s[1]+s[3]>s[2]&&s[2]+s[3]>s[1])return 1;11     return 0;12 }13 int sa()14 {15     int i,j;16     int s;17     for(i=1;i<5;i++)18     for(j=i+1;j<5;j++){19         for(s=1;s<5;s++){20             if(s==i||s==j)continue;21             if(dp[i]+dp[j]==dp[s]){cout<<"SEGMENT"<
        
         >dp[i];}52     sk();53     return 0;54 }