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=3005;
const ll mod=1000000007;
const ll half=500000004;
int n,m;
ll dp[N][N];
ll a[N];
struct node{
	int x,y;ll w;
	node(int x,int y,ll w):x(x),y(y),w(w){}
};
vector<node> v;
int main(){
	int i,j;
	cin>>n>>m;
	for(i=1;i<=n;i++) cin>>a[i];
	for(i=1;i<=n;i++){
		for(j=1;j<=n;j++)
			if(a[i]>a[j]) dp[i][j]=1;
	}
	ll po2=1;
	for(i=1;i<=m;i++) po2=po2*2%mod;
	while(m--){
		v.clear();
		int x,y;cin>>x>>y;
		for(i=1;i<=n;i++){
			if(i!=x&&i!=y){
				v.push_back(node(i,x,dp[i][y]*half%mod));
				v.push_back(node(x,i,dp[y][i]*half%mod));
				v.push_back(node(i,y,dp[i][x]*half%mod));
				v.push_back(node(y,i,dp[x][i]*half%mod));
			}
		}
		v.push_back(node(x,y,dp[y][x]*half%mod));
		v.push_back(node(y,x,dp[x][y]*half%mod));
		for(auto p:v)
			dp[p.x][p.y]=(dp[p.x][p.y]*half+p.w)%mod;
	}
	ll ans=0;
	for(i=1;i<=n;i++){
		for(j=i+1;j<=n;j++)
			ans=(ans+dp[i][j]*po2)%mod;
	}
	cout<<ans;
	return 0;
}