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;

#define MOD 1000000007
#define HALF 500000004

int n, m, x[3005];
long long f[3005][3005], ans, mul = 1;

int ta, tb;

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		scanf("%d", x + i);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			f[i][j] = x[i] < x[j];
	while (m--) {
		mul = (mul << 1) % MOD;
		scanf("%d%d", &ta, &tb);
		ta--;
		tb--;
		f[ta][tb] = f[tb][ta] = HALF * (f[ta][tb] + f[tb][ta]) % MOD;
		for (int i = 0; i < n; i++)
			if (i != ta && i != tb) {
				f[i][ta] = f[i][tb] = HALF * (f[i][ta] + f[i][tb]) % MOD;
				f[ta][i] = f[tb][i] = HALF * (f[ta][i] + f[tb][i]) % MOD;
			}
	}
	for (int i = 0; i < n; i++)
		for (int j = i + 1; j < n; j++)
			ans += f[j][i];
	printf("%lld\n", ans % MOD * mul % MOD);
	return 0;
}