#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
const int nmax=200005;
vector<int> v[nmax];
int arb[4*nmax],best[nmax],b[nmax],cost[nmax];
int n,m,i,j,num,x,y,st,dr,ans,poz,value;
void query(int nod,int l,int r)
{
    if(st<=l&&r<=dr)
    {
        ans=min(ans,arb[nod]);
        return;
    }
    int m=(l+r)/2;
    if(st<=m) query(2*nod,l,m);
    if(m<dr) query(2*nod+1,m+1,r);
}
void update(int nod,int l,int r)
{
    if(l==r)
    {
        arb[nod]=min(arb[nod],value);
        return;
    }
    int m=(l+r)/2;
    if(poz<=m) update(2*nod,l,m);
    else update(2*nod+1,m+1,r);
    arb[nod]=min(arb[2*nod],arb[2*nod+1]);
}
int main()
{
    //freopen("data.in","r",stdin);
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>b[i];
        if(b[i]==0)
        {
            num++;
            cost[i]=-1;
        }
        else cost[i]=1;
    }
    for(i=1;i<=4*n;i++)
        arb[i]=(1<<30);
    cin>>m;
    for(i=1;i<=m;i++)
    {
        cin>>x>>y;
        v[x].push_back(y);
    }
    for(i=1;i<=n;i++)
        best[i]=(1<<30);
    for(i=1;i<=n;i++)
    {
          for(j=0;j<v[i].size();j++)
        {
            st=max(i-1,1);dr=v[i][j];ans=best[i-1];
            query(1,1,n);
            if(ans<best[v[i][j]])
            {
                best[v[i][j]]=ans;
                poz=v[i][j];value=best[v[i][j]];
                update(1,1,n);
            }

        }
        best[i]=min(best[i-1]+cost[i],best[i]);

    }
    cout<<best[n]+num;
    return 0;
}

### Prompt

Please formulate 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 <iostream>
#include <fstream>
#include <vector>
using namespace std;
const int nmax=200005;
vector<int> v[nmax];
int arb[4*nmax],best[nmax],b[nmax],cost[nmax];
int n,m,i,j,num,x,y,st,dr,ans,poz,value;
void query(int nod,int l,int r)
{
    if(st<=l&&r<=dr)
    {
        ans=min(ans,arb[nod]);
        return;
    }
    int m=(l+r)/2;
    if(st<=m) query(2*nod,l,m);
    if(m<dr) query(2*nod+1,m+1,r);
}
void update(int nod,int l,int r)
{
    if(l==r)
    {
        arb[nod]=min(arb[nod],value);
        return;
    }
    int m=(l+r)/2;
    if(poz<=m) update(2*nod,l,m);
    else update(2*nod+1,m+1,r);
    arb[nod]=min(arb[2*nod],arb[2*nod+1]);
}
int main()
{
    //freopen("data.in","r",stdin);
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>b[i];
        if(b[i]==0)
        {
            num++;
            cost[i]=-1;
        }
        else cost[i]=1;
    }
    for(i=1;i<=4*n;i++)
        arb[i]=(1<<30);
    cin>>m;
    for(i=1;i<=m;i++)
    {
        cin>>x>>y;
        v[x].push_back(y);
    }
    for(i=1;i<=n;i++)
        best[i]=(1<<30);
    for(i=1;i<=n;i++)
    {
          for(j=0;j<v[i].size();j++)
        {
            st=max(i-1,1);dr=v[i][j];ans=best[i-1];
            query(1,1,n);
            if(ans<best[v[i][j]])
            {
                best[v[i][j]]=ans;
                poz=v[i][j];value=best[v[i][j]];
                update(1,1,n);
            }

        }
        best[i]=min(best[i-1]+cost[i],best[i]);

    }
    cout<<best[n]+num;
    return 0;
}

```