Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#define MN 200000
#define N 262144
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
vector<int> v[MN+5];
int n,m,a[MN+5],num[MN+5],f[MN+5],T[N*2+5];
inline void Renew(int x,int v){for(x+=N+1;T[x]=min(T[x],v),x>>=1;);}
inline int Query(int l,int r)
{
    int res=1e9;
    for(l+=N,r+=N+2;l^r^1;l>>=1,r>>=1)
    {
        if(~l&1) res=min(res,T[l+1]);
        if( r&1) res=min(res,T[r-1]);
    }
    return res;
}
int main()
{
    n=read();
    for(int i=1;i<=n;++i) num[i]=num[i-1]+1-(a[i]=read()),f[i]=1e9;
    m=read();memset(T,63,sizeof(T));
    for(int i=1;i<=m;++i)
    {
        int l=read(),r=read();
        v[l-1].push_back(r);
    }
    for(int i=0;i<n;++i)
    {
        f[i+1]=min(f[i+1],f[i]+a[i+1]);
        for(int j=0;j<v[i].size();++j)
        {
            int V=min(f[i]+num[v[i][j]]-num[i],Query(i,v[i][j])+num[v[i][j]]);
            f[v[i][j]]=min(f[v[i][j]],V);
            Renew(v[i][j],V-num[v[i][j]]);
        }
    }
    printf("%d",f[n]);
    return 0;
}