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;
const int N=3050,mo=1e9+7,P=(mo+1)/2;
inline int gi() {
  char ch=getchar(); int q=0,x=0;
  while(ch<'0'||ch>'9') q=(ch=='-'?1:q),ch=getchar();
  while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int a[N],f[N][N],g[N][N],p[N],t[N];
int main(){
  int n,q,x,y,s=1; cin>>n>>q;
  for (int i=1; i<=n; ++i) a[i]=gi();
  for (int i=1; i<=n; ++i)
    for (int j=1; j<=n; ++j){
      if (a[i]<a[j]) f[i][j]=1;
      if (a[i]>a[j]) g[i][j]=1;
    }
  for (int i=1; i<=q; ++i){
    x=gi(),y=gi(),s=s*2%mo;
    for (int j=1; j<=n; ++j)
      if(j!=x&&j!=y) {
	f[j][x]=f[j][y]=1ll*P*(f[j][x]+f[j][y])%mo;
	f[x][j]=f[y][j]=1ll*P*(f[x][j]+f[y][j])%mo;
      }
    f[x][y]=f[y][x]=1ll*P*(f[x][y]+f[y][x])%mo;
  }
  int ans=0;
  for(int i=1; i<=n; ++i)
    for(int j=i+1; j<=n; ++j)
      ans=(ans+1ll*f[j][i]*s)%mo;
  cout<<ans;
  return 0;
}