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<cstdio>
#include<algorithm>
using namespace std;
int n, Q, w[3010];
long long Mod = 1000000007, D[3010][3010], TA[3010], TB[3010], SA[3010], SB[3010], I2 = Mod/2+1;
int main() {
	int i, j, a, b;
	scanf("%d%d", &n, &Q);
	for (i = 1; i <= n; i++) {
		scanf("%d", &w[i]);
	}
	for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)if (w[i] < w[j])D[i][j] = 1;
	for(int TT=0;TT<Q;TT++) {
		scanf("%d%d", &a, &b);
		for (i = 1; i <= n; i++) {
			TA[i] = D[a][i];
			TB[i] = D[b][i];
			SA[i] = D[i][a];
			SB[i] = D[i][b];
		}
		for (i = 1; i <= n; i++) {
			if (i != a && i != b) {
				D[a][i] = D[b][i] = (TA[i] + TB[i])*I2%Mod;
				D[i][a] = D[i][b] = (SA[i] + SB[i])*I2%Mod;
			}
		}
		D[a][b] = D[b][a] = (TA[b] + TB[a])*I2%Mod;
	}
	long long res = 0;
	for (i = 1; i <= n; i++)for (j = 1; j < i; j++)res = (res + D[i][j]) % Mod;
	while (Q--)res = res * 2 % Mod;
	printf("%lld\n", res);
}