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

template<typename T>
struct Segtree {
    int n;
    T e;
    vector<T> dat;
    typedef function<T(T a, T b)> Func;
    Func f;

    Segtree(int n_input, Func f, T e_input):f(f), e(e_input){
        n = 1;
        while(n < n_input) n <<= 1;
        dat.resize(2*n-1, e);
    }

    void update(int k, T a){
        k += n - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1)/2;
            dat[k] = f(dat[2*k+1], dat[2*k+2]);
        }
    }

    T get(int k){
        return dat[k+n-1];
    }

    T between(int a, int b){
        return query(a, b+1, 0, 0, n);
    }

    T query(int a, int b, int k, int l, int r){
        if(r<=a || b<=l) return e;
        if(a<=l && r<=b) return dat[k];
        T vl = query(a, b, 2*k+1, l, (l+r)/2);
        T vr = query(a, b, 2*k+2, (l+r)/2, r);
        return f(vl, vr);
    }
};

int main(){
    const int MAX = 200001;
    int N, Q, B[MAX], num0 = 0;
    vector<int> LR[MAX];
    cin >> N;
    for(int i=1; i<=N; i++){
        cin >> B[i];
        if(B[i] == 0) num0++;
    }
    cin >> Q;
    for(int i=0; i<Q; i++){
        int l, r;
        cin >> l >> r;
        LR[l].push_back(r);
    }

    Segtree<int> st(N+1, [](int a, int b){return min(a, b);}, 1e9);
    st.update(0, 0);
    for(int i=1; i<=N; i++){
        sort(LR[i].rbegin(), LR[i].rend());
        for(auto r : LR[i]){
            int result = st.between(i-1, r-1);
            if(result < st.get(r)) st.update(r, result);
        }
        int result = st.get(i-1) + (B[i] ? 1 : -1);
        if(result < st.get(i)) st.update(i, result);
    }

    int ans = st.get(N) + num0;
    cout << ans << endl;
    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 <bits/stdc++.h>
using namespace std;

template<typename T>
struct Segtree {
    int n;
    T e;
    vector<T> dat;
    typedef function<T(T a, T b)> Func;
    Func f;

    Segtree(int n_input, Func f, T e_input):f(f), e(e_input){
        n = 1;
        while(n < n_input) n <<= 1;
        dat.resize(2*n-1, e);
    }

    void update(int k, T a){
        k += n - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1)/2;
            dat[k] = f(dat[2*k+1], dat[2*k+2]);
        }
    }

    T get(int k){
        return dat[k+n-1];
    }

    T between(int a, int b){
        return query(a, b+1, 0, 0, n);
    }

    T query(int a, int b, int k, int l, int r){
        if(r<=a || b<=l) return e;
        if(a<=l && r<=b) return dat[k];
        T vl = query(a, b, 2*k+1, l, (l+r)/2);
        T vr = query(a, b, 2*k+2, (l+r)/2, r);
        return f(vl, vr);
    }
};

int main(){
    const int MAX = 200001;
    int N, Q, B[MAX], num0 = 0;
    vector<int> LR[MAX];
    cin >> N;
    for(int i=1; i<=N; i++){
        cin >> B[i];
        if(B[i] == 0) num0++;
    }
    cin >> Q;
    for(int i=0; i<Q; i++){
        int l, r;
        cin >> l >> r;
        LR[l].push_back(r);
    }

    Segtree<int> st(N+1, [](int a, int b){return min(a, b);}, 1e9);
    st.update(0, 0);
    for(int i=1; i<=N; i++){
        sort(LR[i].rbegin(), LR[i].rend());
        for(auto r : LR[i]){
            int result = st.between(i-1, r-1);
            if(result < st.get(r)) st.update(r, result);
        }
        int result = st.get(i-1) + (B[i] ? 1 : -1);
        if(result < st.get(i)) st.update(i, result);
    }

    int ans = st.get(N) + num0;
    cout << ans << endl;
    return 0;
}
```