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

const int maxn=200005;

int n,q;
int a[maxn];

struct node{
    node* c[2];
    int mx1,mx2;
}pool[maxn<<2];int ps;


node* nwnode(){
    node* rt=pool+ps++;
    rt->mx1=rt->mx2=-(1<<20);
    return rt;
}

void pu(node* rt){
    if(rt&&rt->c[0])rt->mx1=max(rt->c[0]->mx1,rt->c[1]->mx1),rt->mx2=max(rt->c[0]->mx2,rt->c[1]->mx2);
}

node* bt(int l,int r){
    int m=(l+r)>>1;
    node* u=nwnode();
    if(l!=r){
        u->c[0]=bt(l,m);
        u->c[1]=bt(m+1,r);
    }
    return u;
}

void chg(node* rt,int cl,int cr,int i,int v){
    int cm=(cl+cr)>>1;
    if(cl==cr){rt->mx1=max(rt->mx1,v);rt->mx2=max(rt->mx2,v-a[cl]);return;}
    else if(i<=cm)chg(rt->c[0],cl,cm,i,v);
    else chg(rt->c[1],cm+1,cr,i,v);
    pu(rt);
}

int qry(node* rt,int cl,int cr,int l,int r,bool x){
    int cm=(cl+cr)>>1;
    if(cl==l&&cr==r)return x?rt->mx1:rt->mx2;
    else if(r<=cm)return qry(rt->c[0],cl,cm,l,r,x);
    else if(l>cm)return qry(rt->c[1],cm+1,cr,l,r,x);
    else return max(qry(rt->c[0],cl,cm,l,cm,x),qry(rt->c[1],cm+1,cr,cm+1,r,x));
}

pair<int,int> p[maxn];

int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(a[i]==0)a[i]=-1;
        else ans++;
        a[i]+=a[i-1];
        //cerr<<a[i]<<endl;
    }
    cin>>q;
    node* rt=bt(0,n);
    chg(rt,0,n,0,0);
    for(int i=1;i<=q;i++)cin>>p[i].first>>p[i].second;
    sort(p+1,p+1+q);
    for(int i=1;i<=q;i++){
        int el=p[i].first,er=p[i].second;
        int h=qry(rt,0,n,0,el-1,1)+a[er]-a[el-1];
        int g=qry(rt,0,n,el-1,er,0)+a[er];
        chg(rt,0,n,er,max(h,g));
    }
    ans-=qry(rt,0,n,0,n,1);
    cout<<ans;
    return 0;
}
### Prompt

Generate a CPP solution 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;
//dengyaotriangle!

const int maxn=200005;

int n,q;
int a[maxn];

struct node{
    node* c[2];
    int mx1,mx2;
}pool[maxn<<2];int ps;


node* nwnode(){
    node* rt=pool+ps++;
    rt->mx1=rt->mx2=-(1<<20);
    return rt;
}

void pu(node* rt){
    if(rt&&rt->c[0])rt->mx1=max(rt->c[0]->mx1,rt->c[1]->mx1),rt->mx2=max(rt->c[0]->mx2,rt->c[1]->mx2);
}

node* bt(int l,int r){
    int m=(l+r)>>1;
    node* u=nwnode();
    if(l!=r){
        u->c[0]=bt(l,m);
        u->c[1]=bt(m+1,r);
    }
    return u;
}

void chg(node* rt,int cl,int cr,int i,int v){
    int cm=(cl+cr)>>1;
    if(cl==cr){rt->mx1=max(rt->mx1,v);rt->mx2=max(rt->mx2,v-a[cl]);return;}
    else if(i<=cm)chg(rt->c[0],cl,cm,i,v);
    else chg(rt->c[1],cm+1,cr,i,v);
    pu(rt);
}

int qry(node* rt,int cl,int cr,int l,int r,bool x){
    int cm=(cl+cr)>>1;
    if(cl==l&&cr==r)return x?rt->mx1:rt->mx2;
    else if(r<=cm)return qry(rt->c[0],cl,cm,l,r,x);
    else if(l>cm)return qry(rt->c[1],cm+1,cr,l,r,x);
    else return max(qry(rt->c[0],cl,cm,l,cm,x),qry(rt->c[1],cm+1,cr,cm+1,r,x));
}

pair<int,int> p[maxn];

int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(a[i]==0)a[i]=-1;
        else ans++;
        a[i]+=a[i-1];
        //cerr<<a[i]<<endl;
    }
    cin>>q;
    node* rt=bt(0,n);
    chg(rt,0,n,0,0);
    for(int i=1;i<=q;i++)cin>>p[i].first>>p[i].second;
    sort(p+1,p+1+q);
    for(int i=1;i<=q;i++){
        int el=p[i].first,er=p[i].second;
        int h=qry(rt,0,n,0,el-1,1)+a[er]-a[el-1];
        int g=qry(rt,0,n,el-1,er,0)+a[er];
        chg(rt,0,n,er,max(h,g));
    }
    ans-=qry(rt,0,n,0,n,1);
    cout<<ans;
    return 0;
}
```