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;
const int INF = 1011110000, MOD = 1000000007;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
const double EPS = 1e-6;
int n, m;
struct Seg {
  int l, r, x;
  Seg(int _l = 0, int _r = 0, int _x = 0) : l(_l), r(_r), x(_x) {}
  bool operator<(const Seg &that) const { return l < that.l; }
};
struct SegmentTree {
  int n;
  long long e[801111], mk[801111];

 public:
  void init(int _n) {
    n = _n;
    memset(e, 0, sizeof(e));
    memset(mk, 0, sizeof(mk));
  }
  void add(int l, int r, long long x) { add(l, r, x, 0, n, 0); }
  long long sum(int l, int r) { return sum(l, r, 0, n, 0); }

 private:
  void add(int l, int r, long long x, int tl, int tr, int k) {
    if (tr <= l || r <= tl) return;
    if (l <= tl && tr <= r) {
      mk[k] += x;
      e[k] += x * (tr - tl);
      return;
    }
    add(l, r, x, tl, (tl + tr) / 2, k * 2 + 1);
    add(l, r, x, (tl + tr) / 2, tr, k * 2 + 2);
    e[k] = e[k * 2 + 1] + e[k * 2 + 2] + mk[k] * (tr - tl);
  }
  long long sum(int l, int r, int tl, int tr, int k) {
    if (tr <= l || r <= tl) return 0;
    if (l <= tl && tr <= r) {
      return e[k];
    }
    return sum(l, r, tl, (tl + tr) / 2, k * 2 + 1) +
           sum(l, r, (tl + tr) / 2, tr, k * 2 + 2) +
           mk[k] * (min(r, tr) - max(l, tl));
  }
} segt;
bool input() {
  scanf("%d%d", &n, &m);
  return 1;
}
void solve() {
  set<Seg> st;
  segt.init(n);
  for (int i = 0; i < n; i++) {
    st.insert(Seg(i, i + 1, i + 1));
  }
  while (m--) {
    int typ, l, r, x;
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%d%d%d", &l, &r, &x);
      l--;
      Seg s(l, r, x);
      while (1) {
        auto it = st.lower_bound(s);
        if (it == st.end()) break;
        if (it->l < s.r) {
          if (it->r <= s.r) {
            segt.add(it->l, it->r, abs(it->x - x));
            st.erase(it);
          } else {
            segt.add(it->l, r, abs(it->x - x));
            Seg ns = *it;
            ns.l = r;
            st.erase(it);
            st.insert(ns);
          }
        } else
          break;
      }
      auto it = st.lower_bound(s);
      if (it != st.begin()) {
        it--;
        if (l < it->r) {
          if (it->r <= r) {
            segt.add(l, it->r, abs(it->x - x));
            Seg ns = *it;
            ns.r = l;
            st.erase(it);
            st.insert(ns);
          } else {
            segt.add(l, r, abs(it->x - x));
            st.insert(Seg(r, it->r, it->x));
            Seg ns = *it;
            ns.r = l;
            st.erase(it);
            st.insert(ns);
          }
        }
      }
      st.insert(s);
    } else {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", segt.sum(l - 1, r));
    }
  }
}
int main() {
  int zz = 1;
  while (zz--) {
    input();
    solve();
  }
  return 0;
}
