#include <bits/stdc++.h>
using namespace std;
const int MaxN = 200003,INF=0x7f7f7f7f;
int n,b[MaxN],pre[MaxN],q;
vector<int> e[MaxN];

struct Segtree{
    int mn[MaxN*4];
    void modify(int p,int x,int l,int r,int ll,int rr) {
        if (ll<=l && r<=rr) {
            mn[p]=min(mn[p],x);
            return;
        }
        int mid=(l+r)/2;
        if (ll<=mid) modify(p*2,x,l,mid,ll,rr);
        if (rr>mid) modify(p*2+1,x,mid+1,r,ll,rr);
    }
    void modify(int l,int r,int x) {modify(1,x,0,n,l,r);}
    int query(int p,int l,int r,int ll,int rr) {
        int ret=mn[p];
        if (ll<=l && r<=rr) return ret;
        int mid=(l+r)/2;
        if (ll<=mid) ret=min(ret,query(p*2,l,mid,ll,rr));
        if (rr>mid) ret=min(ret,query(p*2+1,mid+1,r,ll,rr));
        return ret;
    }
    int query(int l,int r) {return query(1,0,n,l,r);}
    int query(int x) {return query(1,0,n,x,x);}
    void clear() {memset(mn,0x7f,sizeof mn);}
}sg;

int dp[MaxN];

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;++i) scanf("%d",b+i),pre[i]=pre[i-1]+1-b[i];
    scanf("%d",&q);
    for (int i=1;i<=q;++i) {
        int l,r;
        scanf("%d%d",&l,&r);
        e[r].push_back(l);
    }
    sg.clear();
    for (int i=1;i<=n;++i) {
        sg.modify(i-1,i-1,dp[i-1]-pre[i-1]);
        for (int j:e[i]) sg.modify(j,i,sg.query(j-1));
        dp[i]=min(dp[i-1]+b[i],sg.query(i)+pre[i]);
    }
    printf("%d\n",dp[n]);
    return 0;
}
### Prompt

Please create a solution in cpp to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 200003,INF=0x7f7f7f7f;
int n,b[MaxN],pre[MaxN],q;
vector<int> e[MaxN];

struct Segtree{
    int mn[MaxN*4];
    void modify(int p,int x,int l,int r,int ll,int rr) {
        if (ll<=l && r<=rr) {
            mn[p]=min(mn[p],x);
            return;
        }
        int mid=(l+r)/2;
        if (ll<=mid) modify(p*2,x,l,mid,ll,rr);
        if (rr>mid) modify(p*2+1,x,mid+1,r,ll,rr);
    }
    void modify(int l,int r,int x) {modify(1,x,0,n,l,r);}
    int query(int p,int l,int r,int ll,int rr) {
        int ret=mn[p];
        if (ll<=l && r<=rr) return ret;
        int mid=(l+r)/2;
        if (ll<=mid) ret=min(ret,query(p*2,l,mid,ll,rr));
        if (rr>mid) ret=min(ret,query(p*2+1,mid+1,r,ll,rr));
        return ret;
    }
    int query(int l,int r) {return query(1,0,n,l,r);}
    int query(int x) {return query(1,0,n,x,x);}
    void clear() {memset(mn,0x7f,sizeof mn);}
}sg;

int dp[MaxN];

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;++i) scanf("%d",b+i),pre[i]=pre[i-1]+1-b[i];
    scanf("%d",&q);
    for (int i=1;i<=q;++i) {
        int l,r;
        scanf("%d%d",&l,&r);
        e[r].push_back(l);
    }
    sg.clear();
    for (int i=1;i<=n;++i) {
        sg.modify(i-1,i-1,dp[i-1]-pre[i-1]);
        for (int j:e[i]) sg.modify(j,i,sg.query(j-1));
        dp[i]=min(dp[i-1]+b[i],sg.query(i)+pre[i]);
    }
    printf("%d\n",dp[n]);
    return 0;
}
```