//看看会不会爆int!数组会不会少了一维！

//取物问题一定要小心先手胜利的条件

#include <bits/stdc++.h>

using namespace std;

#pragma comment(linker,"/STACK:102400000,102400000")

#define LL long long

#define ALL(a) a.begin(), a.end()

#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define haha printf("haha\n")

const int maxn = 1e5 + ;

///block表示块的大小，num表示一共分成多少个块

///belong表示这个数在哪个块里面，l和r表示左右边界

/*

查询操作的话，如果x和y是在同一个块里面，就暴力。

如果不是在一个块里面的话，先暴力[x, r[belong[x]]]，然后再暴力块

然后再暴力[l[belong[y]], y]

*/

int block, num, l[maxn], r[maxn], belong[maxn], n, q;

LL Max[maxn], a[maxn];

void build(){

    block = sqrt(n); num = n / block;

    if (n % block) num++;

    for (int i = ; i <= num; i++)

        l[i] = (i - ) * block + , r[i] = i * block;

    r[num] = n;

    for (int i = ; i <= n; i++){

        belong[i] = (i - ) / block + ;

    }

}

void update(int x, int y){

    a[x] += y * 1LL;

    Max[belong[x]] = max(Max[belong[x]], a[x]);

}

int query(int x, int y){

    LL ans = ;

    if (belong[x] == belong[y]){

        for (int i = x; i <= y; i++)

            ans = max(ans, a[i]);

        return ans;

    }

    for (int i = x; i <= r[belong[x]]; i++){

        ans = max(ans, a[i]);

    }

    for (int i = belong[x] + ; i <= belong[y] - ; i++)

        ans = max(ans, Max[i]);

    for (int i = l[belong[y]]; i <= y; i++)

        ans = max(ans, a[i]);

    return ans;

}

int main(){

    cin >> n >> q;

    build();

    for (int i = ; i <= q; i++){

        int op, x, y;

        scanf("%d%d%d", &op, &x, &y);

        if (op == ) update(x, y);

        else printf("%d\n", query(x, y));

    }

    return ;

}