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:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll INF = LLONG_MAX;

class LazySegmentTree {
public:
    int n;
    vector<ll> nodes, lazy;
    LazySegmentTree(int n_) {
        n = 1; while (n_ > n) n <<= 1;
        nodes.resize(n*2-1, 0);
        lazy.resize(n*2-1, 0);
    }
    void eval(int k) {
        if ( ! lazy[k]) return;
        nodes[k] += lazy[k];
        if (k < n-1) {
            lazy[k*2+1] += lazy[k];
            lazy[k*2+2] += lazy[k];
        }
        lazy[k] = 0;
    }
    void add(int l, int r, ll x) {
        add(l, r, 0, n, 0, x);
    }
    void add(int l, int r, int tl, int tr, int k, ll x) {
        eval(k);
        if (tr <= l || r <= tl) return;
        else if (l <= tl && tr <= r) {
            nodes[k] += x;
            if (k < n-1) {
                lazy[k*2+1] += x;
                lazy[k*2+2] += x;
            }
            return;
        }
        add(l, r, tl, (tl+tr)/2, k*2+1, x);
        add(l, r, (tl+tr)/2, tr, k*2+2, x);
        nodes[k] = min(nodes[k*2+1], nodes[k*2+2]);
    }
    ll find(int l, int r) {
        return find(l, r, 0, n, 0);
    }
    ll find(int l, int r, int tl, int tr, int k) {
        eval(k);
        if (tr <= l || r <= tl) return INF;
        else if (l <= tl && tr <= r) return nodes[k];
        return min(find(l, r, tl, (tl+tr)/2, k*2+1), find(l, r, (tl+tr)/2, tr, k*2+2));
    }
};

int main() {
    int n, q;
    cin >> n >> q;
    LazySegmentTree seg(n);
    for (int i=0; i<q; ++i) {
        int a, s, t, x;
        cin >> a;
        if (a == 0) {
            cin >> s >> t >> x;
            seg.add(s, t+1, x);
        } else {
            cin >> s >> t;
            cout << seg.find(s, t+1) << endl;
        }
    }
    return 0;
}