DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

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


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
struct segTree {
  long long n;
  vector<long long> tree, lazy;
  segTree(long long n) : n(n), tree(4 * n, 0), lazy(4 * n, 0) {}
  void propagate(long long node, long long l, long long r) {
    tree[node] += lazy[node] * (r - l + 1);
    if (l != r) lazy[2 * node] += lazy[node], lazy[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
  void treeUpdate(long long node, long long l, long long r, long long u,
                  long long v, long long val) {
    long long mid = (l + r) >> 1;
    propagate(node, l, r);
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      lazy[node] += val;
      propagate(node, l, r);
      return;
    }
    treeUpdate(2 * node, l, mid, u, v, val);
    treeUpdate(2 * node + 1, mid + 1, r, u, v, val);
    tree[node] = tree[2 * node] + tree[2 * node + 1];
  }
  void update(long long l, long long r, long long val) {
    treeUpdate(1, 1, n, l, r, val);
  }
  long long treeQuery(long long node, long long l, long long r, long long u,
                      long long v) {
    long long mid = (l + r) >> 1;
    propagate(node, l, r);
    if (l > v || r < u) return 0;
    if (l >= u && r <= v) return tree[node];
    return treeQuery(2 * node, l, mid, u, v) +
           treeQuery(2 * node + 1, mid + 1, r, u, v);
  }
  long long query(long long l, long long r) { return treeQuery(1, 1, n, l, r); }
};
int main() {
  long long i, j, k, l, m, n, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  while (cin >> n >> m) {
    set<pair<long long, long long> > st;
    segTree tree(n);
    for (i = 1; i <= n + 1; i++) {
      st.insert({i, i});
    }
    for (i = 1; i <= m; i++) {
      long long t;
      scanf("%lld", &t);
      scanf("%lld", &l);
      scanf("%lld", &r);
      if (t == 2) {
        printf("%lld\n", tree.query(l, r));
      } else {
        scanf("%lld", &input);
        set<pair<long long, long long> >::iterator it, jt, kt, mt;
        it = st.lower_bound({l, 0});
        if (it->first != l) {
          jt = it;
          jt--;
          st.insert({l, jt->second});
        }
        it = st.upper_bound({r, LLONG_MAX});
        if (it->first != r + 1) {
          jt = it;
          jt--;
          st.insert({r + 1, jt->second});
        }
        it = st.lower_bound({l, 0});
        jt = st.upper_bound({r, LLONG_MAX});
        for (kt = it; kt != jt; kt = st.erase(kt)) {
          long long a = kt->first;
          mt = kt;
          mt++;
          long long b = mt->first - 1;
          tree.update(a, b, abs(input - kt->second));
        }
        st.insert({l, input});
      }
    }
  }
}
