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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001001001001ll;

int N,Q;
int B[222222];

int acc[222222];
vint qs[222222];

int dp[222222];

struct segtree{
    static const int SEG=1<<18;
    vint dat;
    segtree():dat(SEG*2,INF){}
    void update(int k,int x){
        k+=SEG-1;
        chmin(dat[k],x);
        while(k){
            k=(k-1)/2;
            dat[k]=min(dat[k*2+1],dat[k*2+2]);
        }
    }
    int query(int a,int b,int k=0,int l=0,int r=SEG){
        if(r<=a||b<=l)return INF;
        if(a<=l&&r<=b)return dat[k];
        return min(query(a,b,k*2+1,l,(l+r)/2),query(a,b,k*2+2,(l+r)/2,r));
    }
};

signed main(){
    cin>>N;
    rep(i,N)cin>>B[i];
    rep(i,N)acc[i+1]=acc[i]+1-B[i];
    cin>>Q;
    rep(i,Q){
        int l,r;
        cin>>l>>r;
        l--;
        r--;
        qs[l].pb(r);
    }

    fill_n(dp,222222,INF);
    dp[0]=0;
    segtree seg;

    for(int i=0;i<N;i++){
        for(auto &r:qs[i]){
            int tmp=seg.query(i,r);
            chmin(tmp,dp[i]-acc[i]);
            chmin(dp[r+1],tmp+acc[r+1]);
            seg.update(r,tmp);
        }
        chmin(dp[i+1],dp[i]+B[i]);
    }

    cout<<dp[N]<<endl;
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001001001001ll;

int N,Q;
int B[222222];

int acc[222222];
vint qs[222222];

int dp[222222];

struct segtree{
    static const int SEG=1<<18;
    vint dat;
    segtree():dat(SEG*2,INF){}
    void update(int k,int x){
        k+=SEG-1;
        chmin(dat[k],x);
        while(k){
            k=(k-1)/2;
            dat[k]=min(dat[k*2+1],dat[k*2+2]);
        }
    }
    int query(int a,int b,int k=0,int l=0,int r=SEG){
        if(r<=a||b<=l)return INF;
        if(a<=l&&r<=b)return dat[k];
        return min(query(a,b,k*2+1,l,(l+r)/2),query(a,b,k*2+2,(l+r)/2,r));
    }
};

signed main(){
    cin>>N;
    rep(i,N)cin>>B[i];
    rep(i,N)acc[i+1]=acc[i]+1-B[i];
    cin>>Q;
    rep(i,Q){
        int l,r;
        cin>>l>>r;
        l--;
        r--;
        qs[l].pb(r);
    }

    fill_n(dp,222222,INF);
    dp[0]=0;
    segtree seg;

    for(int i=0;i<N;i++){
        for(auto &r:qs[i]){
            int tmp=seg.query(i,r);
            chmin(tmp,dp[i]-acc[i]);
            chmin(dp[r+1],tmp+acc[r+1]);
            seg.update(r,tmp);
        }
        chmin(dp[i+1],dp[i]+B[i]);
    }

    cout<<dp[N]<<endl;
    return 0;
}

```