The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.

If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.

The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.

Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).

Input

The first line contains two integers n and w (1 ≤ n ≤ 1 000, 1 ≤ w ≤ 10^{9}) — the number of bus stops and the capacity of the bus.

The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} ≤ a_i ≤ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.

Output

Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.

Examples

Input

3 5
2 1 -3


Output

3


Input

2 4
-1 1


Output

4


Input

4 10
2 4 1 2


Output

2

Note

In the first example initially in the bus could be 0, 1 or 2 passengers.

In the second example initially in the bus could be 1, 2, 3 or 4 passengers.

In the third example initially in the bus could be 0 or 1 passenger.
#include <bits/stdc++.h>
using namespace std;
namespace patch {
template <typename T>
std::string to_string(const T& n) {
  std::ostringstream stm;
  stm << n;
  return stm.str();
}
}  // namespace patch
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
bool isAPrime(long long m) {
  for (long long p = 2; p * p <= m; p++) {
    if (m % p == 0) {
      return false;
    }
  }
  return true;
}
bool cmp(int a, int b) { return a > b; }
long long mdf = 1e+9 + 7;
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return (binpow(a, n - 1) * a) % mdf;
  else {
    long long b = binpow(a, n / 2) % mdf;
    return (b * b) % mdf;
  }
}
int count_setbits(int N) {
  int cnt = 0;
  while (N > 0) {
    cnt += (N & 1);
    N = N >> 1;
  }
  return cnt;
}
long long phi(long long n) {
  long long result = n;
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  if (n > 1) result -= result / n;
  return result;
}
bool compare(pair<int, int> a, pair<int, int> b) {
  return a.first < b.first || a.first == b.first && a.second < b.second;
}
void solve() {
  int n, w;
  cin >> n >> w;
  long long sum = 0;
  long long mn = INT_MAX, mx = INT_MIN;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum += x;
    mn = min(sum, mn);
    mx = max(sum, mx);
  }
  mn = min(0ll, mn);
  mx = max(0ll, mx);
  if (-mn > w || mx > w) {
    cout << 0;
    return;
  }
  cout << max(0ll, w - mx + 1 + mn);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q = 1;
  while (q--) solve();
  return 0;
}
