Problem:
RMQ and RAQ

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.

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.

Constraints

$1 ≤ n ≤ 100000$

$1 ≤ q ≤ 100000$

$0 ≤ s ≤ t < n$

$-1000 ≤ x ≤ 1000$

Sample Input 1

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

Sample Output 1

-2
0
1
-1

Solution in C++14 (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define all(c) begin(c),end(c)

template<typename T1, typename T2>
inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }

//改造
typedef long long int ll;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0 //デバッグする時1にしてね
#define dump(x)  cerr << #x << " = " << (x) << endl
#define MOD 1000000007


//ここから編集する
template<typename T, typename E>
struct SegmentTree {
    using F = function<T(T, T)>;
    using G = function<T(T, E)>;
    using H = function<E(E, E)>;
    using P = function<E(E, size_t)>;
    int n;
    F f; //要素と要素のマージ
    G g; //要素に作用素を作用
    H h; //作用素と作用素をマージ
    T ti; //要素の単位元?
    E ei; //作用素の単位元?
    P p; //b個の作用素aをマージした場合
    vector<T> dat;
    vector<E> laz;

    SegmentTree(int n_, F f, G g, H h, T ti, E ei,
                P p = [](E a, size_t b) {
                    b++;
                    return a;
                }) :
            f(f), g(g), h(h), ti(ti), ei(ei), p(p) {
        init(n_);
    }

    void init(int n_) {
        n = 1;
        while (n < n_) n *= 2;
//        dat.assign(2 * n - 1, ti);
        dat.assign(2 * n - 1, 0ll);
        laz.assign(2 * n - 1, ei);
    }

    void build(int n_, vector<T> v) {
        for (int i = 0; i < n_; i++) dat[i + n - 1] = v[i];
        for (int i = n - 2; i >= 0; i--)
            dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);
    }

    inline void eval(int len, int k) {
        if (laz[k] == ei) return;
        if (k * 2 + 1 < n * 2 - 1) {
            laz[k * 2 + 1] = h(laz[k * 2 + 1], laz[k]);
            laz[k * 2 + 2] = h(laz[k * 2 + 2], laz[k]);
        }
        dat[k] = g(dat[k], p(laz[k], len));
        laz[k] = ei;
    }

    T update(int a, int b, E x, int k, int l, int r) {
        eval(r - l, k);
        if (r <= a || b <= l) return dat[k];
        if (a <= l && r <= b) {
            laz[k] = h(laz[k], x);
            return g(dat[k], p(laz[k], r - l));
        }
        return dat[k] = f(update(a, b, x, k * 2 + 1, l, (l + r) / 2),
                          update(a, b, x, k * 2 + 2, (l + r) / 2, r));
    }

    T update(int a, int b, E x) {
        return update(a, b, x, 0, 0, n);
    }

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

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

    /**
     * kをxに更新する
     */
    void update(int k, T x) {
        query(k, k + 1);//evaluate
        k += n - 1;
        dat[k] = x;
        while (k) {
            k = (k - 1) / 2;
            dat[k] = f(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }
};

class Solve {
public:
    void input() {

    }

    void solve() {
        input();

        int n, q;
        cin >> n >> q;
        using T = ll;
        using E = ll;
        auto f = [](T a, T b) { return min(a, b); }; // 要素と要素のマージ
        auto g = [](T a, E b) { return a + b; }; // 要素に作用素を作用
        auto h = [](E a, E b) { return a + b; }; // 作用素と作用素をマージ
        auto p = [](E a, int b) { return a; }; // 作用素を区間加算する場合
        T ti = INFl; //要素の単位元?
        E ei = 0; //作用素の単位元?
        SegmentTree<T, E> treeone(n, f, g, h, ti, ei, p);

        while (q--) {
            int c;
            cin >> c;
            if (c) {
                int s, t;
                cin >> s >> t;
                t++;
                ll val = treeone.query(s, t);
                cout << val << endl;
            } else {
                int s, t;
                ll x;
                cin >> s >> t >> x;
                t++;
                treeone.update(s, t, x);
            }
        }

    }
};


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    Solve().solve();


    return 0;
}