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;

typedef long long LL;
const int MAXN=3111;
const LL Md=1000000007;
const LL Inv2=(Md+1)/2;
LL f[MAXN][MAXN];
LL tf[MAXN][MAXN];
int n, m, a[MAXN];

int main() {
	scanf("%d%d", &n, &m);
	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) {
			f[i][j]=(a[i]>a[j] ? 1 : 0);
		}
	}
	LL wgh=1;
	for(int i=1, x, y;i<=m;++i) {
		wgh=wgh*2ll%Md;
		scanf("%d%d", &x, &y);
		for(int j=1;j<=n;++j) {
			tf[x][j]=(f[y][j]*Inv2%Md+f[x][j]*Inv2%Md)%Md;
			tf[j][x]=(f[j][y]*Inv2%Md+f[j][x]*Inv2%Md)%Md;
			tf[y][j]=(f[y][j]*Inv2%Md+f[x][j]*Inv2%Md)%Md;
			tf[j][y]=(f[j][y]*Inv2%Md+f[j][x]*Inv2%Md)%Md;
		}
		tf[x][y]=(f[x][y]*Inv2%Md+f[y][x]*Inv2%Md)%Md;
		tf[y][x]=(f[y][x]*Inv2%Md+f[x][y]*Inv2%Md)%Md;
		for(int j=1;j<=n;++j) {
			f[x][j]=tf[x][j];
			f[j][x]=tf[j][x];
			f[y][j]=tf[y][j];
			f[j][y]=tf[j][y];
		}
	}
	LL ans=0;
	for(int i=1;i<=n;++i) {
		for(int j=i+1;j<=n;++j) {
			ans=(ans+f[i][j])%Md;
		}
	}
	ans=ans*wgh%Md;
	printf("%lld\n", ans);
	return 0;
}