Problem: Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:

* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.



Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.

Constraints

* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$

Input


$n$ $q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:


0 $s$ $t$ $x$


or


1 $s$ $t$


The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.

Output

For each $find$ query, print the minimum value.

Example

Input

6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5


Output

-2
0
1
-1

Solution (in CPP):

Solution:
//区間加算、区間min

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;


class LazySegmentTree {
public:
    int N;
    vector<ll> node, lazy;
    ll INF = (1LL << 60);

    LazySegmentTree(int _N, ll *a) {
        N = 1;
        while(N < _N) N *= 2;
        node.resize(2 * N - 1, INF);
        lazy.resize(2 * N - 1, 0);

        for(int i = 0; i < _N; i++){
            int k = i + N - 1;
            node[k] = a[i];
        }
        for(int i = N - 2; i >= 0; i--) {
            node[i] = min(node[2 * i + 1], node[2 * i + 2]);
        }
    }

    LazySegmentTree(vector<ll> a) {
        int sz = a.size();
        N = 1;
        while(N < sz) N *= 2;
        node.resize(2 * N - 1, INF);
        lazy.resize(2 * N - 1, 0);

        for(int i = 0; i < sz; i++){
            int k = i + N - 1;
            node[k] = a[i];
        }
        for(int i = N - 2; i >= 0; i--) {
            node[i] = min(node[2 * i + 1], node[2 * i + 2]);
        }
    }

    //[a, b) に c に加算
    void update(int a, int b, ll c) {
        Update(a, b, c, 0, 0, N);
    }

    void Update(int a, int b, ll c, int k, int l, int r) {
        eval(k, l, r);

        //範囲外
        if(b <= l || r <= a) return;

        //完全被覆
        else if(a <= l && r <= b) {
            lazy[k] = c;
            eval(k, l, r);
        }

        //一部被覆
        else {
            Update(a, b, c, 2 * k + 1, l, (l + r) / 2);
            Update(a, b, c, 2 * k + 2, (l + r) / 2, r);
            node[k] = min(node[2 * k + 1], node[2 * k + 2]);
        }

        return;
    }

    //[a, b)のmin
    ll query(int a, int b) {
        return Query(a, b, 0, 0, N);
    }

    ll Query(int a, int b, int k, int l, int r) {
        eval(k, l, r);

        //範囲外
        if(b <= l || r <= a) return INF;

        //完全被覆
        else if(a <= l && r <= b) {
            return node[k];
        }

        //一部被覆
        else {
            ll v1 = Query(a, b, 2 * k + 1, l, (l + r) / 2);
            ll v2 = Query(a, b, 2 * k + 2, (l + r) / 2, r);
            return min(v1, v2);
        }
    }

    void eval(int k, int l, int r) {
        if(lazy[k] != 0) {
            node[k] += lazy[k];
            if(r - l > 1) {
                lazy[2 * k + 1] += lazy[k];
                lazy[2 * k + 2] += lazy[k];
            }
            lazy[k] = 0;
        }
        return;
    }
};

int main(){
    int N, Q;
    cin >> N >> Q;
    ll a[N];
    fill(a, a + N, 0);

    vector<ll> ans;

    LazySegmentTree sg = LazySegmentTree(N, a);
    for(int q = 0; q < Q; q++) {
        int id, s, t, x;
        cin >> id;

        if(id == 0) {
            cin >> s >> t >> x;
            sg.update(s, t + 1, x);
        }
        else {
            cin >> s >> t;
            ans.push_back(sg.query(s, t + 1));
        }
    }

    for(ll val : ans) {
        cout << val << endl;
    }

    return 0;
}