#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
#define all(v) (v).begin(),(v).end()
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
//ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[8] = {0, 0, 1, -1,1,1,-1,-1}, ny[8] = {1, -1, 0, 0,1,-1,1,-1}; //East, West, South, North+
////////////******SOLUTION******\\\\\\\\\\\
const int MAX_N = 300;
ll dp[MAX_N];
int n, q;
vll A;
ll f(int i)
{
if(dp[i] != -1)
return dp[i];
ll maxt = A[i];
ll mint = A[i];
ll ans = 0;
for(int j = i; j <n;j ++)
{
maxt = max(maxt,A[j]);
mint = min(mint,A[j]);
if(j != n-1)
ans = max(ans,maxt-mint+f(j+1));
else
ans = max(ans,maxt-mint);
}
return dp[i] = ans;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> n >> q;
A.assign(n,0);
for(int i = 0; i <n; i ++) cin >> A[i];
while(q--)
{
int l, r;
ll val;
cin >> l>> r >> val;
for(int u = l-1; u < r; u ++)
A[u] += val;
memset(dp,-1,sizeof(dp));
cout << f(0) << '\n';
}
}
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
!!!!! IMPLICIT GRAPH ??????
!!!!! PAY ATTENTION TO THE CONSTRAINTS: DP nD ? BF ? BITMASKING ?
!!!!! SOLVE THE SUBTASKS FIRST: IT'S TOTALLY OK NOT TO SOLVE THE PROBLEM ENTIRELY
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/
