Problem: You are a car race organizer and would like to arrange some races in Linear Kingdom.

Linear Kingdom has n consecutive roads spanning from left to right. The roads are numbered from 1 to n from left to right, thus the roads follow in the order of their numbers' increasing. There will be several races that may be held on these roads. Each race will use a consecutive subset of these roads. Also, each race will pay some amount of money to you if this race is held. No races overlap in time, so some roads can be used in several races.

Unfortunately, some of the roads are in a bad condition and they need repair. Each road has repair costs associated with it, you are required to pay this cost to repair the road. A race can only take place if all the roads used in the race are renovated. Your task is to repair such roads (possibly all or none) that will maximize your profit. Your profit is defined as the total money you get from the races that are held minus the total money you spent to repair the roads. Note that you may decide not to repair any road and gain zero profit.

Print the maximum profit you can gain.

Input

The first line contains two single-space separated integers, n and m (1 ≤ n, m ≤ 2·105), denoting the number of roads and the number of races, respectively.

Then n lines follow, each line will contain a single non-negative integer not exceeding 109 denoting the cost to repair a road. The costs are given in order from road 1 to road n.

Finally, m lines follow. Each line is single-space-separated triplets of integers. Each triplet will be given as lb, ub, and p (1 ≤ lb ≤ ub ≤ n, 1 ≤ p ≤ 109), which means that the race these three integers describe will use all the roads from lb to ub, inclusive, and if it's held you get p.

Output

Print a single integer denoting the maximum possible profit you can gain.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is recommended to use cin, cout stream (also you may use %I64d specificator).

Examples

Input

7 4
3
2
3
2
1
2
3
1 2 5
2 3 5
3 5 3
7 7 5


Output

4


Input

2 1
0
3
1 2 5


Output

2


Input

3 1
10
10
10
1 3 10


Output

0

Note

In the first sample the optimal solution is to repair roads 1, 2, 3, and 7. Three races will take place which nets you 15. The road repair costs 11, hence your profit is 4.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MOD = 1000000007;
const int UNDEF = -1;
const int INF = 1 << 30;
template <typename T>
inline bool chkmax(T &aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T &aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
int rint();
char rch();
long long rlong();
const int mn = 2e5 + 4;
class SegmentTreeFast {
 public:
  long long modifyOperation(long long x, long long y) { return x + y; }
  long long queryOperation(long long l, long long r) { return max(l, r); }
  long long deltaEffectOnSegment(long long delta, int segmentLength) {
    return delta;
  }
  long long initNeutralDelta() { return 0; }
  bool isNeutralDelta(long long delta) { return delta == 0; }
  long long getInitValue(int i) { return 0; }
  long long getNeutralValue() { return 0; }
  long long joinDeltas(long long delta1, long long delta2) {
    return delta1 + delta2;
  }
  long long _cache_neutral_delta;
  long long getNeutralDelta() { return _cache_neutral_delta; }
  long long *value;
  long long *delta;
  int gn;
  long long joinValueWithDelta(long long value, long long delta) {
    if (isNeutralDelta(delta)) return value;
    return modifyOperation(value, delta);
  }
  void pushDelta(int i) {
    int d = 0;
    for (; (i >> d) > 0; d++) {
    }
    for (d -= 2; d >= 0; d--) {
      int x = i >> d;
      value[x >> 1] = joinNodeValueWithDelta(x >> 1, 1 << (d + 1));
      delta[x] = joinDeltas(delta[x], delta[x >> 1]);
      delta[x ^ 1] = joinDeltas(delta[x ^ 1], delta[x >> 1]);
      delta[x >> 1] = getNeutralDelta();
    }
  }
  SegmentTreeFast(int n) {
    _cache_neutral_delta = initNeutralDelta();
    gn = n;
    value = new long long[2 * n];
    for (int i = 0; i < n; i++) {
      value[i + n] = getInitValue(i);
    }
    for (int i = n - 1; i > 0; --i)
      value[i] = queryOperation(value[i << 1], value[i << 1 | 1]);
    delta = new long long[2 * n];
    fill(delta, delta + (2 * n), getNeutralDelta());
  }
  long long joinNodeValueWithDelta(int i, int len) {
    return joinValueWithDelta(value[i], deltaEffectOnSegment(delta[i], len));
  }
  long long query(int from, int to) {
    from += gn;
    to += gn;
    pushDelta(from);
    pushDelta(to);
    long long resl = getNeutralValue();
    long long resr = getNeutralValue();
    bool found = false;
    for (int len = 1; from <= to;
         from = (from + 1) >> 1, to = (to - 1) >> 1, len <<= 1) {
      if ((from & 1) != 0) {
        resl = found ? queryOperation(resl, joinNodeValueWithDelta(from, len))
                     : joinNodeValueWithDelta(from, len);
        found = true;
      }
      if ((to & 1) == 0) {
        resr = found ? queryOperation(joinNodeValueWithDelta(to, len), resr)
                     : joinNodeValueWithDelta(to, len);
        found = true;
      }
    }
    long long res = queryOperation(resl, resr);
    if (!found)
      ;
    return res;
  }
  void modify(int from, int to, long long delta) {
    from += gn;
    to += gn;
    pushDelta(from);
    pushDelta(to);
    int a = from;
    int b = to;
    for (; from <= to; from = (from + 1) >> 1, to = (to - 1) >> 1) {
      if ((from & 1) != 0) {
        this->delta[from] = joinDeltas(this->delta[from], delta);
      }
      if ((to & 1) == 0) {
        this->delta[to] = joinDeltas(this->delta[to], delta);
      }
    }
    for (int i = a, len = 1; i > 1; i >>= 1, len <<= 1) {
      int k = i >> 1;
      value[k] = queryOperation(joinNodeValueWithDelta(k << 1, len),
                                joinNodeValueWithDelta((k << 1) | 1, len));
    }
    for (int i = b, len = 1; i > 1; i >>= 1, len <<= 1) {
      int k = i >> 1;
      value[k] = queryOperation(joinNodeValueWithDelta(k << 1, len),
                                joinNodeValueWithDelta((k << 1) | 1, len));
    }
  }
};
int c[mn];
vector<pair<int, int> > r2lp[mn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n = rint();
  SegmentTreeFast seg = SegmentTreeFast(n + 2);
  int m = rint();
  for (int x = 1; x <= n; x++) c[x] = rint();
  for (int i = 1; i <= m; i++) {
    int l = rint(), r = rint(), p = rint();
    r2lp[r].push_back(make_pair(l, p));
  }
  for (int x = 1; x <= n; x++) {
    {
      long long best = seg.query(0, n + 1);
      seg.modify(x + 1, x + 1, best);
    }
    {
      sort(r2lp[x].begin(), r2lp[x].end());
      for (auto &w : r2lp[x]) {
        int l = w.first, p = w.second;
        seg.modify(0, l, p);
      }
      seg.modify(0, x, -c[x]);
    }
  }
  long long final = seg.query(0, n + 1);
  printf("%lld\n", final);
}
static char stdinBuffer[1024];
static char *stdinDataEnd = stdinBuffer + sizeof(stdinBuffer);
static const char *stdinPos = stdinDataEnd;
void readAhead(size_t amount) {
  size_t remaining = stdinDataEnd - stdinPos;
  if (remaining < amount) {
    memmove(stdinBuffer, stdinPos, remaining);
    size_t sz = fread(stdinBuffer + remaining, 1,
                      sizeof(stdinBuffer) - remaining, stdin);
    stdinPos = stdinBuffer;
    stdinDataEnd = stdinBuffer + remaining + sz;
    if (stdinDataEnd != stdinBuffer + sizeof(stdinBuffer)) *stdinDataEnd = 0;
  }
}
int rint() {
  readAhead(16);
  int x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}
char rch() {
  readAhead(16);
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  char ans = *stdinPos;
  ++stdinPos;
  return ans;
}
long long rlong() {
  readAhead(32);
  long long x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}