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>
typedef unsigned u32;
typedef unsigned long long u64;
const int N=3005;
const u32 P=1e9+7, Half=5e8+4;
int a[N], f[N][N], g[N][N];
int main() {
	int n, q;
	u32 ans=0;
	assert(scanf("%d%d", &n, &q)==2);
	for(int i=1; i<=n; ++i) assert(scanf("%d", a+i)==1);
	for(int i=1; i<=n; ++i) for(int j=1; j<=n; ++j) f[i][j]=a[i]>a[j];
	for(int t=0; t<q; ++t) {
		int x, y;
		assert(scanf("%d%d", &x, &y)==2);
		for(int i=1; i<=n; ++i) {
			g[x][i]=f[x][i];
			g[i][x]=f[i][x];
			g[y][i]=f[y][i];
			g[i][y]=f[i][y];
		}
		for(int i=1; i<=n; ++i)
			if(i!=x&&i!=y) {
				f[x][i]=(u64)Half*(f[x][i]+g[y][i])%P;
				f[i][x]=(u64)Half*(f[i][x]+g[i][y])%P;
				f[y][i]=(u64)Half*(f[y][i]+g[x][i])%P;
				f[i][y]=(u64)Half*(f[i][y]+g[i][x])%P;
			}
		f[x][y]=(u64)Half*(f[x][y]+g[y][x])%P;
		f[y][x]=(u64)Half*(f[y][x]+g[x][y])%P;
	}
	for(int i=1; i<=n; ++i) for(int j=i+1; j<=n; ++j) ans=(ans+f[i][j])%P;
	for(int i=1; i<=q; ++i) ans=ans*2%P;
	printf("%u\n", ans);
	return 0;
}