#include <iostream>
#include <vector>
using namespace std;
int AINT[800005];
int dp[200005];
void build(int nod,int st,int dr)
{
    if(st==dr)
    {
        AINT[nod]=dp[st];
        return ;
    }
    int mid=(st+dr)/2;
    build(nod*2,st,mid);
    build(nod*2+1,mid+1,dr);
    AINT[nod]=min(AINT[2*nod],AINT[2*nod+1]);
}
void u(int nod,int st,int dr,int pos,int val)
{
    if(dr<pos||st>pos)return ;
    if(st==dr){AINT[nod]=val;return ;}
    int mid=(st+dr)/2;
    u(nod*2,st,mid,pos,val);
    u(nod*2+1,mid+1,dr,pos,val);
    AINT[nod]=min(AINT[2*nod],AINT[2*nod+1]);
}
int q(int nod,int st,int dr,int S,int D)
{
    if(dr<S||st>D)return 1<<30;
    if(S<=st&&dr<=D)return AINT[nod];
    int mid=(st+dr)/2;
    return min(q(nod*2,st,mid,S,D),q(nod*2+1,mid+1,dr,S,D));
}
int N,Q;
vector<int> V[200005];
int B[200005],rez;
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
        {cin>>B[i];rez+=!B[i];}
    cin>>Q;
    for(int i=1;i<=Q;i++)
    {
        int x,y;
        cin>>x>>y;
        V[x].push_back(y);
    }
    for(int i=1;i<=N+1;i++)dp[i]=1<<30;
    build(1,0,N+1);
    for(int i=0;i<N;i++)
    {
        vector<int> updates;
        for(auto it:V[i+1])
        {
            dp[it+1]=min(dp[it+1],q(1,0,N+1,i,it+1));
            updates.push_back(it+1);
        }
        dp[i+1]=min(dp[i],dp[i+1])+(B[i+1] ? 1:-1);updates.push_back(i+1);
        dp[i]=1<<30;updates.push_back(i);
        for(auto it:updates)
            u(1,0,N+1,it,dp[it]);
    }
    rez+=min(dp[N+1],dp[N]);
    cout<<rez;
    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 <vector>
using namespace std;
int AINT[800005];
int dp[200005];
void build(int nod,int st,int dr)
{
    if(st==dr)
    {
        AINT[nod]=dp[st];
        return ;
    }
    int mid=(st+dr)/2;
    build(nod*2,st,mid);
    build(nod*2+1,mid+1,dr);
    AINT[nod]=min(AINT[2*nod],AINT[2*nod+1]);
}
void u(int nod,int st,int dr,int pos,int val)
{
    if(dr<pos||st>pos)return ;
    if(st==dr){AINT[nod]=val;return ;}
    int mid=(st+dr)/2;
    u(nod*2,st,mid,pos,val);
    u(nod*2+1,mid+1,dr,pos,val);
    AINT[nod]=min(AINT[2*nod],AINT[2*nod+1]);
}
int q(int nod,int st,int dr,int S,int D)
{
    if(dr<S||st>D)return 1<<30;
    if(S<=st&&dr<=D)return AINT[nod];
    int mid=(st+dr)/2;
    return min(q(nod*2,st,mid,S,D),q(nod*2+1,mid+1,dr,S,D));
}
int N,Q;
vector<int> V[200005];
int B[200005],rez;
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
        {cin>>B[i];rez+=!B[i];}
    cin>>Q;
    for(int i=1;i<=Q;i++)
    {
        int x,y;
        cin>>x>>y;
        V[x].push_back(y);
    }
    for(int i=1;i<=N+1;i++)dp[i]=1<<30;
    build(1,0,N+1);
    for(int i=0;i<N;i++)
    {
        vector<int> updates;
        for(auto it:V[i+1])
        {
            dp[it+1]=min(dp[it+1],q(1,0,N+1,i,it+1));
            updates.push_back(it+1);
        }
        dp[i+1]=min(dp[i],dp[i+1])+(B[i+1] ? 1:-1);updates.push_back(i+1);
        dp[i]=1<<30;updates.push_back(i);
        for(auto it:updates)
            u(1,0,N+1,it,dp[it]);
    }
    rez+=min(dp[N+1],dp[N]);
    cout<<rez;
    return 0;
}

```