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 N=3e3+100,mod=1e9+7,inv2=(mod+1)>>1;
int n,Q,a[N],f[N][N],fx[N],gx[N],Ans;
void Inc(int &x,int y) {x+=y, x=x>=mod?x-mod:x;}
int qpow(LL x,int y)
{
	LL ret=1;
	while(y)
	{
		if(y&1) ret=ret*x%mod;
		x=x*x%mod, y>>=1;
	}
	return ret;
}

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)
			f[i][j]=(a[i]>a[j])?1:0;
	for(int q=1;q<=Q;++q)
	{
		int x,y; scanf("%d%d",&x,&y);
		if(x == y) continue;
		for(int i=1;i<=n;++i)
			fx[i]=1ll*inv2*(f[x][i]+f[y][i])%mod,
			gx[i]=1ll*inv2*(f[i][x]+f[i][y])%mod;
		fx[y]=1ll*inv2*(f[x][y]+f[y][x])%mod;
		for(int i=1;i<=n;++i)
			f[x][i]=fx[i], f[i][x]=gx[i],
			f[y][i]=fx[i], f[i][y]=gx[i];
		f[x][y]=f[y][x]=fx[y], f[x][x]=f[y][y]=0;
	}
	for(int i=1;i<=n;++i)
		for(int j=i+1;j<=n;++j) Inc(Ans,f[i][j]);
	Ans=1ll*Ans*qpow(2,Q)%mod; printf("%d",Ans);
	return 0;
}