Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int N = 3005 , mod = 1e9 + 7 ;
long long inv2 = 5e8 + 4 , mul = 1;
int n , q , a[N] , p[N][N] , ans = 0;

int main() {
    scanf("%d %d" , &n , &q);
    for(int i = 1 ; i <= n ; ++i) scanf("%d" , a + i);
    for(int i = 1 ; i <= n ; ++i)
        for(int j = 1 ; j <= n ; ++j) p[i][j] = (a[i] < a[j]);
    while(q--) {
        mul = (mul << 1) % mod;
        int a , b;
        scanf("%d %d" , &a , &b);
        for(int i = 1 ; i <= n ; ++i)
            if(i != a && i != b) {
                p[i][b] = p[i][a] = (p[i][a] * inv2 + p[i][b] * inv2) % mod;
                p[b][i] = p[a][i] = (p[b][i] * inv2 + p[a][i] * inv2) % mod;
            }
        p[a][b] = p[b][a] = (p[a][b] * inv2 + p[b][a] * inv2) % mod;
    }
    for(int i = 1 ; i <= n ; ++i)
        for(int j = i + 1 ; j <= n ; ++j) ans = (ans + mul * p[j][i]) % mod;
    printf("%d\n" , ans);
    return 0;
}