Problem: Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.

Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).

Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.

Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.

Input

The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.

The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.

Output

If it is impossible to obtain a tank with exactly V ml of water, print NO. 

Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format: 

Each line has to contain 3 numbers denoting a compressed operation: "cnt x y" (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y. 

The number of these lines must not exceed N + 5.

Examples

Input

2 3 5
2 3


Output

YES
1 2 1


Input

2 3 4
2 3


Output

NO


Input

5 2 0
1 3 5 7 9


Output

YES
2 2 1
3 3 1
4 4 1
5 5 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
using namespace std;
class pa3 {
 public:
  long long x, y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  double x;
  int y, z, w;
  pa4(double x = 0, int y = 0, int z = 0, int w = 0) : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  int x, y;
  pa2(int x = 0, int y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
class Point {
 public:
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double a) { return Point(x * a, y * a); }
  Point operator/(double a) { return Point(x / a, y / a); }
  double absv() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  bool operator<(const Point &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const Point &p) const {
    return fabs(x - p.x) < (1e-10) && fabs(y - p.y) < (1e-10);
  }
};
struct Segment {
  Point p1, p2;
};
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
bool parareru(Point a, Point b, Point c, Point d) {
  return abs(cross(a - b, d - c)) < (1e-10);
}
double distance_ls_p(Point a, Point b, Point c) {
  if (dot(b - a, c - a) < (1e-10)) return (c - a).absv();
  if (dot(a - b, c - b) < (1e-10)) return (c - b).absv();
  return abs(cross(b - a, c - a)) / (b - a).absv();
}
bool is_intersected_ls(Segment a, Segment b) {
  if (a.p1 == b.p1 || a.p2 == b.p1 || a.p1 == b.p2 || a.p2 == b.p2)
    return false;
  if (parareru((a.p2), (a.p1), (a.p1), (b.p2)) &&
      parareru((a.p2), (a.p1), (a.p1), (b.p1))) {
    if (dot(a.p1 - b.p1, a.p1 - b.p2) < (1e-10)) return true;
    if (dot(a.p2 - b.p1, a.p2 - b.p2) < (1e-10)) return true;
    if (dot(a.p1 - b.p1, a.p2 - b.p1) < (1e-10)) return true;
    if (dot(a.p1 - b.p2, a.p2 - b.p2) < (1e-10)) return true;
    return false;
  } else
    return (cross(a.p2 - a.p1, b.p1 - a.p1) * cross(a.p2 - a.p1, b.p2 - a.p1) <
            (1e-10)) &&
           (cross(b.p2 - b.p1, a.p1 - b.p1) * cross(b.p2 - b.p1, a.p2 - b.p1) <
            (1e-10));
}
double segment_dis(Segment a, Segment b) {
  if (is_intersected_ls(a, b)) return 0;
  double r = distance_ls_p(a.p1, a.p2, b.p1);
  r = min(r, distance_ls_p(a.p1, a.p2, b.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p1));
  return r;
}
Point intersection_ls(Segment a, Segment b) {
  Point ba = b.p2 - b.p1;
  double d1 = abs(cross(ba, a.p1 - b.p1));
  double d2 = abs(cross(ba, a.p2 - b.p1));
  double t = d1 / (d1 + d2);
  return a.p1 + (a.p2 - a.p1) * t;
}
string itos(int i) {
  ostringstream s;
  s << i;
  return s.str();
}
int gcd(int v, int b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
double distans(double x1, double y1, double x2, double y2) {
  double rr = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
  return sqrt(rr);
}
int mod;
int beki(int wa, int rr, int warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  if (rr % 2 == 1) return (beki(wa, rr - 1, warukazu) * wa) % warukazu;
  int zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long a[6000];
bool er[5100][5100] = {0};
bool dp[5100][5100] = {0};
int mae[5100][5100];
vector<pa3> ans;
vector<int> ve, ve2;
int q[6000] = {0};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k, v;
  cin >> n >> k >> v;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
    if (a[i] == v) {
      cout << "YES" << endl;
      return 0;
    }
  }
  if (sum < v) {
    cout << "NO" << endl;
    return 0;
  }
  if (v % k == 0) {
    cout << "YES" << endl;
    for (int i = 2; i <= n; i++) {
      int kai = a[i] / k + !!(a[i] % k);
      if (kai) cout << kai << " " << i << " " << 1 << endl;
      a[1] += a[i];
    }
    if (v / k) cout << v / k << " 1 2" << endl;
    return 0;
  }
  dp[0][0] = 1;
  mae[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < k; j++) {
      if (dp[i][j]) {
        dp[i + 1][j] = 1;
        mae[i + 1][j] = i * 10000;
        mae[i + 1][j] += j;
        er[i + 1][j] = 0;
        dp[i + 1][(j + a[i + 1]) % k] = 1;
        mae[i + 1][(j + a[i + 1]) % k] = i * 10000;
        mae[i + 1][(j + a[i + 1]) % k] += j;
        er[i + 1][(j + a[i + 1]) % k] = 1;
      }
    }
  if (dp[n][v % k] == 0) {
    cout << "NO" << endl;
    return 0;
  }
  int cn1 = n, cn2 = v % k;
  while (1) {
    if (er[cn1][cn2]) ve.push_back(cn1);
    int ccn1 = cn1, ccn2 = cn2;
    cn1 = mae[ccn1][ccn2] / 10000;
    cn2 = mae[ccn1][ccn2] % 10000;
    if (cn1 == 0) break;
  }
  if (ve.size() == n) {
    cout << "YES" << endl;
    for (int i = 2; i <= n; i++) {
      long long kai = a[i] / k + !!(a[i] % k);
      cout << kai << " " << i << " 1" << endl;
      a[1] += a[i];
      a[i] = 0;
    }
    int f = a[1] - v;
    if (f / k > 0) cout << f / k << " 1 2" << endl;
    return 0;
  }
  if (ve.size() == 0) {
    cout << "NO" << endl;
    return 0;
  }
  for (auto v : ve) {
    q[v] = 1;
  }
  int l = ve.size();
  for (int i = 1; i < l; i++) {
    long long kai = a[ve[i]] / k + !!(a[ve[i]] % k);
    if (kai) ans.push_back((pa3){kai, ve[i], ve[0]});
    a[ve[0]] += a[ve[i]];
    a[ve[i]] = 0;
  }
  for (int i = 1; i <= n; i++)
    if (q[i] == 0) {
      ve2.push_back(i);
    }
  l = ve2.size();
  for (int i = 1; i < l; i++) {
    long long kai = a[ve2[i]] / k + !!(a[ve2[i]] % k);
    if (kai) ans.push_back((pa3){kai, ve2[i], ve2[0]});
    a[ve2[0]] += a[ve2[i]];
    a[ve2[i]] = 0;
  }
  if (a[ve2[0]] >= k) {
    ans.push_back((pa3){a[ve2[0]] / k, ve2[0], ve[0]});
    a[ve[0]] += a[ve2[0]] / k * k;
    a[ve2[0]] %= k;
  }
  if (a[ve[0]] < v) {
    cout << "NO" << endl;
    return 0;
  }
  int g = a[ve[0]] - v;
  if (g / k > 0) ans.push_back((pa3){g / k, ve[0], ve2[0]});
  cout << "YES" << endl;
  for (auto v : ans) cout << v.x << " " << v.y << " " << v.z << endl;
  return 0;
}