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 <stdio.h>

const long long mod = 1000000007;
const long long inv2 = (mod + 1) / 2;

int N,Q,A[3030],C[3030][3030];

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++) C[i][j] = A[i] > A[j];

	long long mul = 1;
	for (int q=1;q<=Q;q++){
		int x,y; scanf ("%d %d",&x,&y);
		for (int i=1;i<=N;i++) if (i != x && i != y){
			int p = C[x][i] + C[y][i];
			int q = C[i][x] + C[i][y];
			C[x][i] = C[y][i] = p * inv2 % mod;
			C[i][x] = C[i][y] = q * inv2 % mod;
		}
		int r = C[x][y] + C[y][x];
		C[x][y] = C[y][x] = r * inv2 % mod;
		mul = mul * 2 % mod;
	}

	long long ans = 0;
	for (int i=1;i<=N;i++) for (int j=i+1;j<=N;j++) ans = (ans + C[i][j] * mul) % mod;
	printf ("%lld\n",ans);
	
	return 0;
}