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<bits/stdc++.h>
using namespace std;

int n,q,sz,w[200005],s[200005],dp[200005];
vector<int> g[200005];
int dat[800005];

void upd(int l,int r,int x){
    l+=sz;r+=sz;
    while(l<=r){
        dat[l]=min(dat[l],x);
        dat[r]=min(dat[r],x);
        l=(l+1)>>1;r=(r-1)>>1;
    }
}

int qry(int x){
    int r=0x3f3f3f3f;
    x+=sz;
    while(x){
        r=min(r,dat[x]);
        x>>=1;
    }
    return r;
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    memset(dat,0x3f,sizeof(dat));
    sz=1;while(sz<n)sz<<=1;
    for(int i=1;i<=n;i++){
        cin>>w[i];
        s[i]=s[i-1]+1-w[i];
    }
    cin>>q;
    while(q--){
        int a,b;
        cin>>a>>b;
        g[b].push_back(a);
    }
    for(int i=1;i<=n;i++){
        upd(i-1,i-1,dp[i-1]-s[i-1]);
        for(int &j:g[i]){
            upd(j,i,qry(j-1));
        }
        dp[i]=min(dp[i-1]+w[i],qry(i)+s[i]);
    }
    cout<<dp[n]<<endl;

    return 0;
}