Vova has taken his summer practice this year and now he should write a report on how it went.

Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.

Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages. 

However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.

Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.

Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.

Input

The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).

The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.

The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.

Output

Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.

Otherwise print "NO".

Examples

Input


2 2
5 5
2 2


Output


YES


Input


2 2
5 6
2 2


Output


NO


Input


4 1
4 1 10 1
3 2 10 1


Output


YES

Note

In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'): 

  * page 1: "TTFTTFT" 
  * page 2: "TFTTFTT" 



That way all blocks of tables have length 2.

In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
#include <bits/stdc++.h>
using namespace std;
constexpr long double m_pi = 3.1415926535897932L;
constexpr long long MOD = 1000000007;
constexpr long long INF = 1LL << 62;
constexpr long double EPS = 1e-10;
template <typename T>
using vector2 = vector<vector<T>>;
template <typename T>
using vector3 = vector<vector2<T>>;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  int to, rev;
  long long cap;
  Edge(int _to, long long _cap, int _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, int from, int to, long long cap,
              bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size() + (from == to)));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (long long(i) = (long long)(0); i < (long long)(used.size()); i++)
      used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    negative[i] = false;
  d[s] = 0;
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (long long(i) = (long long)(0); i < (long long)(G[a.second].size());
         i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i].resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(d[i].size()); j++) {
      d[i][j] = ((i != j) ? INF : 0);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
      chmin(d[i][G[i][j].to], G[i][j].cap);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G.size()); j++) {
      for (long long(k) = (long long)(0); k < (long long)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<long long>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (long long(i) = (long long)(0); i < (long long)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class Lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  Lca() {}
  Lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  long long size(int x) { return -data[root(x)]; }
  long long numSet() { return num; }
};
template <typename T, typename F>
class SegmentTree {
 private:
  T identity;
  F merge;
  long long n;
  vector<T> dat;

 public:
  SegmentTree(F f, T id, vector<T> v) : merge(f), identity(id) {
    int _n = v.size();
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
    for (long long(i) = (long long)(0); i < (long long)(_n); i++)
      dat[n + i - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
  }
  SegmentTree(F f, T id, int _n) : merge(f), identity(id) {
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
  }
  void set_val(int i, T x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  T query(int l, int r) {
    T left = identity, right = identity;
    l += n - 1;
    r += n - 1;
    while (l < r) {
      if ((l & 1) == 0) left = merge(left, dat[l]);
      if ((r & 1) == 0) right = merge(dat[r - 1], right);
      l = l / 2;
      r = (r - 1) / 2;
    }
    return merge(left, right);
  }
};
template <typename T>
class FenwickTree {
  vector<T> data;
  int n;
  int p;

 public:
  FenwickTree(int n) : n(n) {
    data.resize(n + 1LL, 0);
    p = 1;
    while (p < data.size()) p *= 2;
  }
  T sum(int k) {
    T ret = 0;
    for (; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }
  T sum(int a, int b) { return sum(b) - sum(a); }
  void add(int k, T x) {
    for (++k; k <= n; k += k & -k) data[k] += x;
  }
  int lower_bound(long long w) {
    if (w <= 0) return -1;
    int x = 0;
    for (int k = p / 2; k > 0; k /= 2) {
      if (x + k <= n && data[x + k] < w) w -= data[x + k], x += k;
    }
    return x;
  }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
void prime_factorization(long long n, vector<pair<long long, long long>>& ret) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back({i, 0});
      while (n % i == 0) {
        n /= i;
        ret[ret.size() - 1].second++;
      }
    }
  }
  if (n != 1) ret.push_back({n, 1});
}
inline long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
inline long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> fact_inv;
  long long mod;
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    if (n < mod) return ((fact[n] * fact_inv[r] % mod) * fact_inv[n - r]) % mod;
    long long ret = 1;
    while (n || r) {
      long long _n = n % mod, _r = r % mod;
      n /= mod;
      r /= mod;
      (ret *= nCr(_n, _r)) %= mod;
    }
    return ret;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * fact_inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long _n, long long _mod) {
    mod = _mod;
    long long n = min(_n + 1, mod);
    fact.resize(n);
    fact[0] = 1;
    for (long long(i) = (long long)(0); i < (long long)(n - 1); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    fact_inv.resize(n);
    fact_inv[n - 1] = mod_inv(fact[n - 1], mod);
    for (int i = n - 1; i > 0; i--) {
      fact_inv[i - 1] = fact_inv[i] * i % mod;
    }
  }
};
long long popcount(long long x) {
  x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
  x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
  x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
  x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
  x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
  x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> x(n), y(n);
  for (long long(i) = (long long)(0); i < (long long)(n); i++) cin >> x[i];
  for (long long(i) = (long long)(0); i < (long long)(n); i++) cin >> y[i];
  long long px = 0, py = 0;
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    long long mx = (py <= k ? k : k - px);
    long long my = (px <= k ? k : k - py);
    long long tx = INF, ty = INF;
    if (mx > 0) {
      long long c = min(x[i] - 1, y[i]);
      if (c * k >= y[i]) {
        long long tmp = max(1LL, x[i] - mx - (c - 1) * k);
        chmin(tx, tmp);
      }
      c = min(x[i], y[i]);
      if ((c - 1) * k + mx >= x[i]) {
        long long tmp = max(1LL, y[i] - (c - 1) * k);
        chmin(ty, tmp);
      }
    }
    if (my > 0) {
      long long c = min(x[i], y[i] - 1);
      if (c * k >= x[i]) {
        long long tmp = max(1LL, y[i] - my - (c - 1) * k);
        chmin(ty, tmp);
      }
      c = min(x[i], y[i]);
      if ((c - 1) * k + my >= y[i]) {
        long long tmp = max(1LL, x[i] - (c - 1) * k);
        chmin(tx, tmp);
      }
    }
    px = tx;
    py = ty;
  }
  if (px <= k || py <= k)
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
  return 0;
}
