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 int long long
const int mod=1e9+7;
int dp[3005][3005],n,q,val[3005],now[3005][3005],inv2;
int qpow(int x,int y)
{
	int res=1;
	while(y)
	{
		if(y&1)res=res*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return res;
}
signed main()
{
	inv2=qpow(2,mod-2);
	scanf("%lld%lld",&n,&q);
	for(int i=1;i<=n;++i)scanf("%lld",&val[i]);
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=n;++j)
		{
			dp[i][j]=(val[i]<val[j]?1:0);
		}
	}
	int mx=qpow(2,q);
	while(q--)
	{
		int x,y;
		scanf("%lld%lld",&x,&y);
		for(int i=1;i<=n;++i)
		{
			now[x][i]=(dp[x][i]+dp[y][i])%mod*inv2%mod;
			now[i][x]=(dp[i][x]+dp[i][y])%mod*inv2%mod;
			now[y][i]=now[x][i];
			now[i][y]=now[i][x];
		}
		now[x][y]=now[y][x]=(dp[x][y]+dp[y][x])*inv2%mod;
		for(int i=1;i<=n;++i)
    {
			dp[x][i]=now[x][i];dp[i][x]=now[i][x];
			dp[y][i]=now[y][i];dp[i][y]=now[i][y];
		}
		swap(val[x],val[y]);
	}
	int res=0;
	for(int i=1;i<=n;++i)
	{
		for(int j=i+1;j<=n;++j)
		{
			res+=mx*dp[j][i]%mod;
			if(res>=mod)res-=mod;
		}
	}
	cout<<res<<endl;
  return 0;
}